Anova

1

Analysis Of Variance

To test Ho: μ1=μ2 when δ1 and δ2 are identical, we use two samples ttest. Now if we wish to test, Ho: μ1=μ2=------------μk. Then we cannot apply two samples t-test on each pair, then we have kC2 tests which is very difficult to manage. And the situation will be incorrect because the overall significance level (1-α) is greatly increased as the number of tests increases. Overall significance level= 1-(αper test)no. of tests Suppose we wish to test k=4 population means simultaneously then we have 4C2 = 6 two sample t-tests suppose α = 0.05. Then Overall significance level = 1-(αper test)no. of tests = 1- (0.05)6 = 0.99999 Facing such situations R.A Fisher introduced the technique of ANOVA in which total variation is partitioned into its components part, each of which have different source of variation. Thus in ANOVA we compare different estimates of variance by using F- distribution to test the equality of several population means. 2: Assumption of One-Way ANOVA: (1). (2). (3). (4). 3:

The samples observations will be randomly selected. The sampled population should be normal. All the k-populations having identical variance. All effects are additive. Xij = μ + Ti + tij ONE-WAY ANOVA

When each observation is classified according to a single criterion of classification, then we use One-Way ANOVA.

Let Xij be the ith observation of the jth sample (treatment), then it can be arranged as; Samples (Treatments) Observations 1 2…………j……………..…k 1 Xij X12 ………X1j……………..X1k 2 X21 X22………..X2j…………….X2k . . . . . . . . . . . . . . . . . . . . i Xi1 Xi2………..Xij………………Xik . . . . . . . . . . . . . . . . . . . . r Xr1 Xr2……….Xrj………………Xrk Total T.1 T.2……….T.j………………T.k Mean x .1 x .2 ……….. x .j………..….. x .k r

k

∑∑

The term

(xi- x )2 is called total variation. Now total variation can

i =1 j =1

be partitioned into its components parts as; r

k

∑∑ i =1 j =1 r

(xij -

x

)

r

2

x

i =1 j =1

i =1 j =1

..)2 + (xij -

k

= ∑∑ [( x ij -

x

i =1 j =1 r

∑∑

[xij -

x

…xj-

x ]2

k

= ∑∑ [(xi r

=

k

..)2 + (xij -

k

= ∑∑ ( i =1 j =1

r

xj

-

x

x j)

..)2 +

2

]

x j)

2

+2( x ij +

k

∑∑ i =1 j =1

x

)(xij r

(xij -

2

x j)

x j)]

k

+ 2 ∑∑ ( x ii =1 j =1

x

..)(xij -

As i and j are independent, then the cross product term will be zero r

k

∑∑ i =1 j =1

(xij -

x

..)2 = x ∑ ( x j -

SSTotal = SSTreatment + SSError

r

x

)2 +

k

∑∑ i =1 j =1

(xij +

2

x j)

x i)

S.O.V Treatment Error Total F=MST / MSE

ANOVA Table d.f SS MS k-1 SST SST / k-1 =MST n-k SSE SSE / n-k=MSE n-1 SSTotal

F F=MST / MSE

~ F(k-1,n-k)

Computation Formula: C.F = (T..)2 / n SSTotal = ∑∑ X2ij – C.F SST = ∑ Tj2 / r – C.F SSE = SSTotal – SST Q 20.8 1. Ho: μ1=μ2=μ3=μ4 VS H1: μ1ǂμ2ǂμ3ǂμ4 2. The significance level: α=0.05 3. Test statistic: If Ho is true then F= MST/ MSE 4.

T.j T.j2

~ F (3,16)

Computation: Machine Number 1 2 3 64 41 64 39 48 57 65 41 76 46 49 72 63 57 64 277 236 334 76729 55696 111556

∑∑ Xij2= (64)2 +………+(47)2= 61937

4 45 51 55 48 47 246 6051 6

1093 304497

C.F = (T..)2/n = (1093)2/20 = 59732.45 SSTotal = ∑∑Xij2 – C.F = 61937 – 59732.45 = 2204.55 SST = ∑T.j2/r – C.F = 304497/5 – 59732.45 = 116695 SSE = SSTotal – SST = 2204.55 – 116.95 = 1037.6

ANOVA Table S.O.V d.f SS B/w treatment 3 1166.95 Error 16 1037.6 Total 9 2204.55 As F = MST / MSE = 6.00

MS F 388.98 F = 6 64.85

Reject Ho: when F > F0.05(3,16) = 3.24 Remarks: As F-calculated value falls in the rejection region, so we cannot accept our Ho at α=0.05.

5. 6.

Q20.13 1. 2. 3.

4.

Ho: μ1=μ2=μ3 VS H1: μ1ǂμ2ǂμ3 The significance level: α=0.05 Test statistic: if Ho is true, then F = MST / MSE ~ F (2,10) Computation: Methods 1 2 3 47.2 50.1 49.1 49.8 49.3 53.2 48.5 51.5 51.2 48.7 50.9 52.8 52.3 T.j 194.2 201.8 258.6 T.j2 3772364 40723.24 66873.9 6

∑Xij2 = (47.2)2 +……….+(52.8)2 = 33001

654.6

C.F = (T..)2/n = (654.6)2/13 = 32961.63 SSTotal = ∑∑Xij2 – C.F = 33001-32961 = 39.37 SST = T.12/r1 + T.22/r2 + T.32/r3 – C.F = 37713.64/4 + 40723.34/4 + 66873.96/5 – 32961.63 = 22.382 SSE = SSTotal – SST = 39.37 – 22.382 = 16.988

ANOVA Table S.O.V B/w treatment Error Total

d.f 2 10 12

SS 22.382 16.988 39.37

MS F 11.191 F=6.60 1.6988

As F = MST / MSE = 6.60 5. 6.

4:

Reject Ho: when F > F0.05(2,10) = 4.10 Remarks: As F-calculated value falls in the rejection region, so we cannot accept our Ho at α=0.05 TWO WAY ANOVA

When each observation is classified according to two criterion of classification, then we use two way ANOVA, one criterion is taken on row side and second criterion on column side. Let Xij denoted the ith observation in the ith row and jth column, and then it can be arranged as;

Observations 1 2 . . . . i . . . . r Total Mean

r

The term

Treatment 1 2…………j……………..…c X12 ………X1j……………..X X22………..X2j…………….X2 . . . . . . . . . . . . Xi2………..Xij………………X . . . . . . . . . . . . Xr2……….Xrj………………Xr T.2……….T.j………………T

1 Xij X21 . . . . Xi1 . . . . Xr1 T.1 x

.1

x

.

x

Total T.1 T.2 . . . . T.i . . . . T.r

Mean x .1 x .2 . . . . x .i . . . . x .r x

.2 ……….. x .j………..…..

..

c

∑∑

(xij -

i =1 j =1

x

..)2 is called variation.

Now total variation can be partitioned as; r

c

∑∑ i =1 j =1 r

(xij -

..)

x

r

2

=

c

∑∑ i =1 j =1

(xij -

x

..+ x j +

xj

-

x i+ x

.. x ..)2

c

= ∑∑ [( x j -

x

..) + ( x i -

x

..) + ( x ij -

x

..)2 + ( x i -

x

..)2 + ( x ij -

i =1 j =1 r

xj

-

xi

+

x

..)]2

c

= ∑∑ [( x j i =1 j =1 r

= r∑ ( j =1

x j- x

S.O.V b/w columns

2

..) + c

r

∑ i1

d.f c-1

( x i-

x

x j- x i+ x

..)2 + ∑∑ ( x ij -

..)2

x j - x i+ x

ANOVA Table SS MS SSC SSC / c-1 = MSC

..)2

F F1= MSC / MSE

b/w rows Error Total

r-1 (r-1)(c-1) n-1

SSR SSE SSTotal

SSR / r-1 = MSR SSE / (r-1)(c-1) = MSE F2= MSR / MSE

F1= MSC / MSE

~

F2= MSR / MSE

~ F(r-1, (r-1)(c-1))

F(c-1, (r-1)(c-1))

R.. C.F = (T..)2/n SSTotal = ∑∑ Xij2 – C.F SSC = ∑ T.i2 /r – C.F SSR = ∑ T.j2 /c – C.F SSE = SSTotal – SSC – SSR

Q20.21

1.

H`o: μ1=μ2=μ3=μ4 vs H`1: μ1ǂμ2ǂμ3ǂμ4 H``o: μ1=μ2=μ3 vs H``1: μ1ǂμ2ǂμ3 2. The significance level α=0.05 3. Test statistic: if H`o and H``o are true, then F1 = MSC / MSE ~ F (3,6) F2 = MSR / MSE ~ F (2,6) 4. Computation: B1 B2 B3 B4 R1 46.5 62 41 45 R2 47.5 41.5 22 31.5 R3 50 40 25.5 28.5 T.j 144 143.5 88.5 105 2 T.j 2073 20592.2 7832.2 11025 6 5 5

T.i 194.5 142.5 144 481 60185. 5

T.i2 37830.25 20306.25 20736 78872.5

∑∑ Xij2 = (46.5)2+-----------+(28.5)2 = 207929.5 C.F = (T..)2/n = (481)2/12 = 19280.08 SSTotal = ∑∑ Xij2 – C.F = 207929.5 – 19280.08 = 1449.42 SSC = ∑T.j2 / r – C.F 60185.5 / 3 – 19280.08 = 781.75 SSR = ∑ T.i2/C – C.F 78872.5 / 4 – 19280.08 = 438.045 SSE = SSTotal – SSC – SSR = 1449.42 – 781.75 – 438.045 = 229.625 ANOVA Table S.O.V d.f SS B/w columns 3 781.75 B/w Rows 2 438.045 Error 6 229.625 Total 11 1449.42

5. 6.

MS 260.58 219.02 38.271

F F1= 6.8 F2= 5.72

Reject H`o: when F1 > F0.05(3,6) = 4.76 Reject H``o: when F2 > F0.05(2,6) = 5.14 Remarks: As F1 and F2 calculated value falls in the rejection region, so we cannot accept our H`o and H``o at α=0.05

Q20.22 1. H`o: μ1=μ2=μ3=μ4=μ5 vs H`1: μ1ǂμ2ǂμ3ǂμ4ǂμ5 H``o: μ1=μ2=μ3=μ4 vs H``1: μ1ǂμ2ǂμ3ǂμ4 2. The significance level: α = 0.05 3. Test statistic: if H`o and H``o are true, then

F1 = MSC / MSE

~ F (4,12)

F2 = MSR / MSE

~ F (3,12)

4. Computation:

1 2 3 4 T.j T.j2

1 1.9 2.5 1.7 2.1 8.2 67.2 4

2 2.2 1.9 1.9 1.8 7.8 60.8 4

3 2.6 2.3 2.2 2.5 9.6 92.16

4 1.8 2.6 2.0 2.3 8.7 75.69

5 2.1 2.2 2.1 2.4 8.8 77.44

C.F = (T..)2/n = (43.1)2/20 = 92.88 ∑Xij2 = (1.9)2 +-----------+(2.4)2 = 94.31 SSTotal = ∑Xij2 – C.F = 94.31 – 92.88 = 1.43 SSC = ∑T.j2/r – C.F = 373.37/4 – 92.88 = 0.4625 SSR = ∑T.i2/c – C.F = 465.83/5 – 92.88 = 0.286 SSE = SSTotal – SSC – SSR = 1.43 – 0.4625 – 0.286 = 0.6815 S.O.V d.f B/w columns 4 B/w Rows Error

3 12

ANOVA Table SS MS 0.462 0.1156 5 0.286 0.0915 0.681 0.0568 5

Total 19 5. Reject H`o: when F1 > F0.05(4,12) = 3.26 Reject H``o: when F2 > F0.05(3,12) = 3.49

F F1= 2.035 F2= 1.667

T.i 10.6 11.5 9.9 11.1 43.1 373.37

T.i2 112.36 132.25 98.01 123.21 465.83

6. Remarks: As F1 and F2 calculated value falls in the acceptance region,

so we cannot reject our H`o and H``o at α=0.05 5. Fisher Least Significant Difference Test (LSD): In case of ANOVA, when H0 is rejected all we conclude that the population means are significant, but this conclusion might be not satisfy the experimenter. The experimenter further wants to know which pair of means is significant and which are not? To deal with such situation Sir R.A Fisher introduced a significant difference test known as Least Significant Difference test (LSD), which is the generation of Student’s T-distribution given by; 2 M SE r

LSD= t α / 2 (v)

Where v = Error d.f When r1ǂ r2 ǂ r3 ǂ………..ǂ rk

Then r=

K

1 1 + + 1 .........+ rk1 r1 r 2 r 3

t α / 2 (v) = t0.025(16) = 2.120

20.8

LSD = t α / 2 (v) = 2.120

x

.1 = 227/5

x

.3 = 336/5 = 66.8

2 M SE r

2x 64.8 5 5

= 10.8

= 55.4

x

.2

x

x

.4

.2 = 236/5 = 47.2 x

.4 = 246/5 = 49.2

x

.1

x

.3

47.2 49.2 55.4 _____________ _______________________ 20.13

t α / 2 (v) = t0.025(10) = 2.228 r=

K

=

1 1 + + 1 .........+ rk1 r1 r 2 r 3

= 2.228

3

= 4.28

1 1 1 + + 4 4 5

2 M SE r

LSD= t α / 2 (v)

x

2 (1.6988) 4.28

= 1.98

.1 = 194.2 / 4 = 48.55 .2 = 201.8 / 4 = 50.45 .3 = 258.6 / 5 = 51.72 x

x

x

.1

x

48.55

.2

50.45

x

.3

51.72

_____________ ________________________ 20.21

For Column wise: t α / 2 (v) = t0.025(6) = 2.447 LSD= t α / 2 (v) = 2.447 x x

66.8

2 M SE r 2 ( 38.271) 3

= 12.36

.1 = 144 / 3 = 48 .2 = 143.5 / 3 = 47.83 .3 = 88.5 / 3 = 29.5 .4 = 105 / 3 = 35 x

x

x

.3

x

.4

x

29.5 35 ______________

.2

x

.1

47.83 48 _______________

For Row wise: 2 M SE c

LSD= t α / 2 (v)

2 ( 38.271) 4

= 2.447 x

= 10.70

.1 = 194.5 / 4 = 48.625 .2 = 142.5 / 4 = 35.625 .3 = 144 / 4 = 36 x

x

x

.2

x

.3

x

35.625 36 _____________

6.

.1

48.625

Duncan’s Multiple Range Test (DMRT): In case of ANOVA when null hypothesis is rejected all we conclude that the population means are significant, but this conclusion might not be satisfy the experimenter. The experimenter further wants to know which pair of means is significant and which are not. To deal with such situation D.B Duncan proposed a multiple comparison test known as DMRT given by; Rp = q α (p,v)

M SE r

Where P = 2,3,4,…….. And When

v = Error d.f

r1ǂ r2 ǂ r3 ǂ………..ǂ rk

r=

K

1 1 + + 1 .........+ rk1 r1 r 2 r 3

To carry out the test, we arrange the sample means in ascending order of magnitude and compare with various Duncan’s values. x

20.8

.1= 55.4 M SE r =

p

q0.05 (p, 16)

2 3 4

3.00 3.15 3.23

1

64.8 5 5

.2= 47.2

x

4 vs 1:

x

4 vs 2:

x

4 vs 3:

x

3 vs 1:

x

3 vs 2:

x

2 vs 1:

x

3

.4 .3 – .3 – .3 – .1 – .1 – .4 –

M SE r

Rp = q0.05 (p, 16)

2 x

.3=66.8

= 3.601

10.803 11.343 11.631

.2

x

x

x

x x

x x

4

.1

x

.3

.2 = 19.63* > (R ) .4 = 17.6* > (R ) .1 = 11.1* > (R ) .2 = 8.2 < (R ) .4 = 6.2 < (R ) .2 = 2 < (R )

x x

(R2) (R3) (R4)

4

3

2

3

2 2

Note: * means significant. 20.21

For Column wise: x x

.1 = 144 / 3 = 48 .3 = 88.5 / 3 = 29.5

x x

.2 = 143.5 / 3 = 47.83 .4 = 105 / 3 = 35

x

.4= 49.2

M SE r =

38.271 3

= 3.571

p

q0.05 (p, 6)

2 3 4

3.46 3.58 3.64

1 x

12.355 12.784 12.998

2

.3

29.5

x

x

4 vs 2:

x

4 vs 3:

x

3 vs 1:

x

3 vs 2:

x

2 vs 1:

x

(R2) (R3) (R4)

3

.4

x

35

4 vs 1:

4

.2

x

47.5

.1 – .1 – .1 – .2 – .2 – .4 –

M SE r

Rp = q0.05 (p, 6)

48

.3 = 18.5* > (R ) .4 = 13* > (R ) .2 = 0.5 > (R ) .3 = 18* < (R ) .4 = 12 < (R ) .3 = 5 < (R )

x x

x x

x x

.1

4

3

2

3

2

2

Note: * means significant. For Row wise:

x

.1 = 194.5 / 4 = 48.625 .2 = 142.5 / 4 = 35.625 .3 = 144 / 4 = 36 x

x

M SE = c

38.27 1 4 =

3.093

p

q0.05 (p, 6)

2 3

3.46 3.58

10.70 11.07

1

3

x

35.625

36

3 vs 1:

x

3 vs 2:

x

2 vs 1:

x

(R2) (R3)

2

.2

x

M SE c

Rp = q0.05 (p, 6)

.3

.1 – .2 = 13* .1 – .3= 12.625* .3 – .2 = 0.375

x

.1

48.625

x

<

(R3 )

x

<

(R2 )

x

<

(R2 )

Note: * means significant. --------------------------------------------